Chapter 5

Metal-Casting Processes andEquipment; Heat Treatment

Questions

5.1 Describe the characteristics of (1) an alloy, (2)pearlite, (3) austenite, (4) martensite, and (5)cementite.

(a) Alloy: composed of two or more elements,at least one element is a metal. The al-loy may be a solid solution or it may formintermetallic compounds.

(b) Pearlite: a two-phase aggregate consistingof alternating lamellae of ferrite and ce-mentite; the closer the pearlite spacing oflamellae, the harder the steel.

(c) Austenite: also called gamma iron, it hasa fcc crystal structure which allows for agreater solubility of carbon in the crys-tal lattice. This structure also possesses ahigh ductility, which increases the steel’sformability.

(d) Martensite: forms by quenching austen-ite. It has a bct (body-centered tetrag-onal) structure, and the carbon atoms ininterstitial positions impart high strength.It is hard and very brittle.

(e) Cementite: also known as iron-carbide(Fe3C), it is a hard and brittle intermetal-lic phase.

5.2 What are the effects of mold materials on fluidflow and heat transfer?

The most important factor is the thermal con-ductivity of the mold material; the higher theconductivity, the higher the heat transfer andthe greater the tendency for the fluid to solid-ify, hence possibly impeding the free flow of themolten metal. Also, the higher the cooling rateof the surfaces of the casting in contact withthe mold, the smaller the grain size and hencethe higher the strength. The type of surfacesdeveloped in the preparation of mold materi-als may also be different. For example, sand-mold surfaces are likely be rougher than thoseof metal molds whose surfaces can be preparedto varying degrees of roughness, including thedirections of roughness (lay).

5.3 How does the shape of graphite in cast iron af-fect its properties?

The shape of graphite in cast irons has the fol-lowing basic forms:

(a) Flakes. Graphite flakes have sharp edgeswhich act as stress raisers in tension.This shape makes cast iron low in tensilestrength and ductility, but it still has highcompressive strength. On the other hand,the flakes also act as vibration dampers,a characteristic important in damping ofmachine-tool bases and other structures.

(b) Nodules. Graphite can form nodules orspheroids when magnesium or cerium is

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added to the melt. This form has in-creased ductility, strength, and shock re-sistance compared to flakes, but the damp-ing ability is reduced.

(c) Clusters. Graphite clusters are much likenodules, except that they form from thebreakdown of white cast iron upon anneal-ing. Clusters have properties that are ba-sically similar to flakes.

(d) Compacted flakes. These are short andthick flakes with rounded edges. This formhas properties that are between nodularand flake graphite.

5.4 Explain the difference between short and longfreezing ranges. How are they determined?Why are they important?

Freezing range is defined by Eq. (5.3) on p. 196in terms of temperature difference. Referringto Fig. 5.6 on p. 197, note that once the phasediagram and the composition is known, we candetermine the freezing range, TL − TS . As de-scribed in Section 5.3.2 on p. 196, the freezingrange has an important influence on the for-mation and size of the mushy zone, and, conse-quently, affects structure-property relationshipsof the casting.

5.5 We know that pouring molten metal at a highrate into a mold has certain disadvantages.Are there any disadvantages to pouring it veryslowly? Explain.

There are disadvantages to pouring metalslowly. Besides the additional time needed formold filling, the liquid metal may solidify orpartially solidify while still in the gating systemor before completely filling the mold, resultingin an incomplete or partial casting. This canhave extremely detrimental effects in a tree ofparts, as in investment casting.

5.6 Why does porosity have detrimental effects onthe mechanical properties of castings? Whichphysical properties are also affected adverselyby porosity?

Pores are, in effect, internal discontinuities thatare prone to cracking and crack propagation.Thus, the toughness of a material will decreaseas a result of porosity. Furthermore, the pres-ence of pores in a piece of metal under tension

indicates that the material around the pores hasto support a greater load than if no pores werepresent; thus, the strength is also lowered. Con-sidering thermal and electrical conductivity, aninternal defect such as a pore decreases both thethermal and electrical conductivity, nting thatair is a very poor conductor.

5.7 A spoked hand wheel is to be cast in gray iron.In order to prevent hot tearing of the spokes,would you insulate the spokes or chill them?Explain.

Referring to Table 5.1 on p. 206, we note that,during solidification, gray iron undergoes an ex-pansion of 2.5%. Although this fact may sug-gest that hot tearing cannot occur, considera-tion must also be given to significant contrac-tion of the spokes during cooling. Since the hot-tearing tendency will be reduced as the strengthincreases, it would thus be advisable to chill thespokes to develop this strength.

5.8 Which of the following considerations are im-portant for a riser to function properly? (1)Have a surface area larger than that of the partbeing cast. (2) Be kept open to atmosphericpressure. (3) Solidify first. Explain.

Both (1) and (3) would result in a situationcontrary to a riser’s purpose. That is, if a risersolidifies first, it cannot feed the mold cavity.However, concerning (2), an open riser has someadvantages over closed risers. Recognizing thatopen risers have the danger of solidifying first,they must be sized properly for proper func-tion. But if the riser is correctly sized so that itremains a reservoir of molten metal to accom-modate part shrinkage during solidification, anopen riser helps exhaust gases from the moldduring pouring, and can thereby eliminate someassociated defects. A so-called blind riser thatis not open to the atmosphere may cause pock-ets of air to be trapped, or increased dissolu-tion of air into the metal, leading to defects inthe cast part. For these reasons, the size andplacement of risers is one of the most difficultchallenges in designing molds.

5.9 Explain why the constant C in Eq. (5.9) de-pends on mold material, metal properties, andtemperature.

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The constant C takes into account various fac-tors such as the thermal conductivity of themold material and external temperature. Forexample, Zircon sand has a higher thermal con-ductivity than basic silica sand, and as a result,a casting in Zircon (of equal volume and surfacearea) will require less time to solidify than thatcast in silica.

5.10 Explain why gray iron undergoes expansion,rather than contraction, during solidification.

As gray cast iron solidifies, a period ofgraphitization occurs during the final stages,which causes an expansion that counteracts theshrinkage of the metal. This results in an over-all expansion.

5.11 How can you tell whether a cavity in a castingare due to porosity or to shrinkage?

Evidence of which type of porosity is present,i.e., gas or shrinkage, can be gained by study-ing the location and shape of the cavity. If theporosity is near the mold surface, core surface,or chaplet surface, it is most likely to be gasporosity. However, if the porosity occurs in anarea considered to be a hot spot in the casting(see Fig. 5.37 on p. 249), it is most likely tobe shrinkage porosity. Furthermore, gas poros-ity has smooth surfaces (much like the holes inSwiss cheese) and is often, though not always,generally spherical in shape. Shrinkage poros-ity has a more textured and jagged surface, andis generally irregular in shape.

5.12 Explain the reasons for hot tearing in castings.

Hot tearing is a result of tensile stresses that de-velop upon contraction during solidification inmolds and cores if they are not sufficiently col-lapsible and/or do not allow movement underthe resulting pressure during shrinkage.

5.13 Would you be concerned about the fact thata portion of an internal chill is left within thecasting? What materials do you think chillsshould be made of, and why?

The fact that a part of the chill remains withinthe casting should be a consideration in the de-sign of parts to be cast. The following factorsare important:

(a) The material from which the chill is madeshould be compatible with the metal be-ing cast (it should have approximatelythe same composition of the metal beingpoured).

(b) The chill must be clean, that is, withoutany lubricant or coating on the surface,because any gas evolved when the moltenmetal contacts the chill may not readilyescape.

(c) The chill may not fuse with the casting,developing regions of weakness or stressconcentration. If these factors are under-stood and provided for, the fact that apiece of the chill remains within the cast-ing is generally of no significant concern.

5.14 Are external chills as effective as internal chills?Explain.

The effectiveness will depend on the location ofthe region to be chilled in the mold. If a regionneeds to be chilled (say, for example, to direc-tionally solidify a casting), an external chill canbe as effective as an internal chill. Often, how-ever, chilling is required at some depth beneaththe surface of a casting to be effective. For thiscondition an internal chill would be more effec-tive.

5.15 Do you think early formation of dendrites in amold can impede the free flow of molten metalinto the mold? Explain.

Consider the solidification of an alloy with avery long freezing range. The mushy zone forthis alloy will also be quite large (see Fig. 5.6).Since the mushy condition consists of interlac-ing dendrites surrounded by liquid, it is appar-ent that this condition will restrict fluid flow,as also confirmed in practice.

5.16 Is there any difference in the tendency forshrinkage void formation for metals with shortfreezing and long freezing ranges, respectively?Explain.

In an alloy with a long freezing range, the pres-ence of a large mushy zone is more likely tooccur, and thus the formation of miocroporos-ity. However, in an alloy with a short freezingrange, the formation of gross shrinkage voids ismore likely to occur.

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5.17 It has long been observed by foundrymen thatlow pouring temperatures (that is, low super-heat) promote equiaxed grains over columnargrains. Also, equiaxed grains become finer asthe pouring temperature decreases. Explain thereasons for these phenomena.

Equiaxed grains are present in castings near themold wall where rapid cooling and solidificationtake place by heat transfer through the rela-tively cool mold. With low pouring tempera-ture, cooling to the solidification temperatureis faster because there is less heat stored in themolten metal. With a high pouring tempera-ture, cooling to the solidification temperatureis slower, especially away from the mold wall.The mold still dissipates heat, but the metal re-mains molten for a longer period of time, thusproducing columnar grains in the direction ofheat conduction. As the pouring temperature isdecreased, equiaxed grains should become finerbecause the cooling is more rapid and largegrains do not have time to form from the moltenmetal.

5.18 What are the reasons for the large variety ofcasting processes that have been developed overthe years?

By the student. There are several acceptableanswers depending on the interpretation of theproblem by the student. Students may ap-proach this as processes that are applicationdriven, material driven, or economics driven.For example, while investment casting is moreexpensive than sand casting, closer dimensionaltolerances and better surface finish are possi-ble. Thus, for certain parts such as barrelsfor handguns, investment casting is preferable.Consider also the differences between the hot-and cold-chamber permanent-mold casting op-erations.

5.19 Why can blind risers be smaller than open-toprisers?

Risers are used as reservoirs for a casting in re-gions where shrinkage is expected to occur, i.e,areas which are the last to solidify. Thus, risersmust be made large enough to ensure that theyare the last to solidify. If a riser solidifies beforethe cavity it is to feed, it is useless. As a result,an open riser in contact with air must be larger

to ensure that it will not solidify first. A blindriser is less prone to this phenomenon, as it isin contact with the mold on all surfaces; thus ablind riser may be made smaller.

5.20 Would you recommend preheating the molds inpermanent-mold casting? Also, would you re-move the casting soon after it has solidified?Explain.

Preheating the mold in permanent-mold castingis advisable in order to reduce the chilling effectof the metal mold which could lead to low metalfluidity and the problems that accompany thiscondition. Also, the molds are heated to re-duce thermal damage which may result fromrepeated contact with the molten metal. Con-sidering casting removal, the casting should beallowed to cool in the mold until there is nodanger of distortion or developing defects dur-ing shakeout. While this may be a very shortperiod of time for small castings, large castingsmay require an hour or more.

5.21 In a sand-casting operation, what factors deter-mine the time at which you would remove thecasting from the mold?

This question is an important one for any cast-ing operation, not just sand casting, becausea decrease in production time will result in adecrease in product cost. Therefore, a castingideally should be removed at the earliest pos-sible time. Factors which affect time are thethermal conductivity of the mold-material andof the cast metal, the thickness and the over-all size of the casting, and the temperature atwhich the metal is being poured.

5.22 Explain why the strength-to-weight ratio of die-cast parts increases with decreasing wall thick-ness.

Because the metal die acts as a heat sink for themolten metal, the metal chills rapidly, develop-ing a fine-grain hard skin with higher strength.As a result, the strength-to-weight ratio of die-cast parts increases with decreasing wall thick-ness.

5.23 We note that the ductility of some cast alloysis very low (see Fig. 5.13). Do you think thisshould be a significant concern in engineeringapplications of castings? Explain.

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The low ductility of some cast alloys should cer-tainly be taken into consideration in the engi-neering application of the casting. The low duc-tility will:

(a) affect properties, such as toughness and fa-tigue,

(b) have a significant influence on furtherprocessing and finishing of the casting,i.e., machining processes, such as milling,drilling, and tapping, and

(c) possibly affect tribological behavior.

It should be noted that many engineering ap-plications do not require high ductility; for ex-ample, when stresses are sufficiently small toensure the material remains elastic and whereimpact loads do not occur.

5.24 The modulus of elasticity of gray iron variessignificantly with its type, such as the ASTMclass. Explain why.

Because the shape, size, and distribution ofthe second-phase (i.e., the graphite flakes) varygreatly for gray cast irons, there is a large cor-responding variation of properties attainable.The elastic modulus, for example, is one prop-erty which is affected by this factor.

5.25 List and explain the considerations involved inselecting pattern materials.

Pattern materials have a number of importantmaterial requirements. Often, they are ma-chined, thus good machinability is a require-ment. The material should be sufficiently stiffto allow good shape development. The materialmust have sufficient wear and corrosion resis-tance so that the pattern has a reasonable life.The economics of the operation is affected alsoby material cost.

5.26 Why is the investment-casting process capableof producing fine surface detail on castings?

The surface detail of the casting depends onthe quality of the pattern surface. In invest-ment casting, for example, the pattern is madeof wax or a thermoplastic poured or injectedinto a metal die with good surface finish. Con-sequently, surface detail of the casting is verygood and can be controlled. Furthermore, the

coating on the pattern (which then becomes themold) consists of very fine silica, thus contribut-ing to the fine surface detail of the cast product.

5.27 Explain why a casting may have a slightly dif-ferent shape than the pattern used to make themold.

After solidification, shrinkage continues untilthe casting cools to room temperature. Also,due to surface tension, the solidifying metalwill, when surface tension is high enough, notfully conform to sharp corners and other intri-cate surface features. Thus, the cast shape willgenerally be slightly different from that of thepattern used.

5.28 Explain why squeeze casting produces partswith better mechanical properties, dimensionalaccuracy, and surface finish than expendable-mold processes.

The squeeze-casting process consists of a com-bination of casting and forging. The pressureapplied to the molten metal by the punch, orupper die, keeps the entrapped gases in solu-tion, and thus porosity is generally not foundin these products. Also, the rapid heat transferresults in a fine microstructure with good me-chanical properties. Due to the applied pres-sure and the type of die used, i.e., metal, gooddimensional accuracy and surface finish are typ-ically found in squeeze-cast parts.

5.29 Why are steels more difficult to cast than castirons?

The primary reason steels are more difficult tocast than cast irons is that they melt at a highertemperature. The high temperatures compli-cate mold material selection, preparation, andtechniques involved for heating and pouring ofthe metal.

5.30 What would you recommend to improve thesurface finish in expendable-mold casting pro-cesses?

One method of improving the surface finish ofcastings is to use what is known as a facingsand, such as Zircon. This is a sand having bet-ter properties (such as permeability and surfacefinish) than bulk sand, but is generally moreexpensive. Thus, facing sand is used as a first

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layer against the pattern, with the rest of themold being made of less expensive (silica) sand.

5.31 You have seen that even though die casting pro-duces thin parts, there is a limit to the mini-mum thickness. Why can’t even thinner partsbe made by this process?

Because of the high thermal conductivity thatmetal dies exhibit, there is a limiting thicknessbelow which the molten metal will solidify pre-maturely before filling the mold cavity. Also,the finite viscosities of the molten metal (whichincreases as it begins to cool) will require higherpressures to force the metal into the narrow pas-sages of the die cavities.

5.32 What differences, if any, would you expect inthe properties of castings made by permanent-mold vs. sand-casting methods?

As described in the text, permanent-moldcastings generally possess better surface fin-ish, closer tolerances, more uniform mechani-cal properties, and more sound thin-walled sec-tions than sand castings. However, sand cast-ings generally can have more intricate shapes,larger overall size, and lower in cost (dependingupon the alloy) than permanent-mold castings.

5.33 Which of the casting processes would be suit-able for making small toys in large numbers?Explain.

This is an open-ended problem, and the stu-dents should give a rationale for their choice.Refer also to Table 5.2 and note that die castingis one of the best processes for this application.The student should refer to the application re-quiring large production runs, so that toolingcost per casting can be low, the sizes possiblein die casting are suitable for such toys, and thedimensional tolerances and surface finish are ac-ceptable.

5.34 Why are allowances provided for in making pat-terns? What do they depend on?

Shrinkage allowances on patterns are correc-tions for the shrinkage that occurs upon solidifi-cation of the casting and its subsequent contrac-tion while cooling to room temperature. Theallowance will therefore depend on the amountof contraction an alloy undergoes.

5.35 Explain the difference in the importance ofdrafts in green-sand casting vs. permanent-mold casting.

Draft is provided to allow the removal of thepattern without damaging the mold. If themold material is sand and has no draft, themold cavity is likely to be damaged upon pat-tern removal, due to the low strength of thesand mold. However, a die made of high-strength steel, which is typical for permanent-mold castings, is not at all likely to be damagedduring the removal of the part; thus smallerdraft angles can be employed.

5.36 Make a list of the mold and die materials usedin the casting processes described in this chap-ter. Under each type of material, list the cast-ing processes that are used, and explain whythese processes are suitable for that particularmold or die material.

This is an open-ended problem, and studentsshould be encouraged to develop an answerbased on the contents of this chapter. An ex-ample of an acceptable answer would, in a briefform, be:

• Sand: Used because of its ability to resistvery high temperatures, availability, andlow cost. Used for sand, shell, expanded-pattern, investment, and ceramic-moldcasting processes.

• Metal: Such as steel or iron. Results inexcellent surface finish and good dimen-sional accuracy. Used for die, slush, pres-sure, centrifugal, and squeeze-casting pro-cesses.

• Graphite: Used for conditions similar tothose for metal molds; however, lowerpressures should be employed for this ma-terial. Used mainly in pressure- andcentrifugal-casting.

• Plaster of paris: Used in plaster-mold cast-ing for the production of relatively smallcomponents, such as fittings and valves.

5.37 Explain why carbon is so effective in impartingstrength to iron in the form of steel.

Carbon has an atomic radius that is about 57%of the iron atom, thus it occupies an interstitialposition in the iron unit cell (see Figs. 3.2 on

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p. 84 and 3.9 on p. 90). However, because itsradius is greater than that of the largest holebetween the Fe atoms, it strains the lattice,thus interfering with dislocation movement andleading to strain hardening. Also, the size ofthe carbon atom allows it to have a high solu-bility in the fcc high-temperature phase of iron(austenite). At low temperatures, the structureis bcc and has a very low solubility of carbonatoms. On quenching, the austenitic structuretransforms to body-centered tetragonal (bct)martensite, which produces high distortion inthe crystal lattice. Because it is higher, thestrength increase is more than by other elementadditions.

5.38 Describe the engineering significance of the ex-istence of a eutectic point in phase diagrams.

The eutectic point corresponds to a compo-sition of an alloy that has a lowest meltingtemperature for that alloy system. The lowmelting temperature associated with a eutec-tic point can, for example, help in controllingthermal damage to parts during joining, as isdone in soldering. (See Section 12.13.3 startingon p. 776).

5.39 Explain the difference between hardness andhardenability.

Hardness represents the material’s resistance toplastic deformation when indented (see Section2.6 starting on p. 51), while hardenability isthe material’s capability to be hardened by heattreatment. (See also Section 5.11.1 starting onp. 236).

5.40 Explain why it may be desirable or necessary forcastings to be subjected to various heat treat-ments.

The morphology of grains in an as-cast struc-ture may not be desirable for commercial appli-cations. Thus, heat treatments, such as quench-ing and tempering (among others), are carriedout to optimize the grain structure of castings.In this manner, the mechanical properties canbe controlled and enhanced.

5.41 Describe the differences between case hardeningand through hardening insofar as engineeringapplications are concerned.

Case hardening is a treatment that hardensonly the surface layer of the part (see Table5.7 on p. 242). The bulk retains its toughness,which allows for blunting of surface cracks asthey propagate inward. Case hardening gener-ally induces compressive residual stresses on thesurface, thus retarding fatigue failure. Throughhardened parts have a high hardness across thewhole part; consequently, a crack could propa-gate easily through the cross section of the part,causing major failure.

5.42 Type metal is a bismuth alloy used to cast typefor printing. Explain why bismuth is ideal forthis process.

When one considers the use of type or for preci-sion castings such as mechanical typewriter im-pressions, one realizes that the type tool musthave extremely high precision and smooth sur-faces. A die casting using most metals wouldhave shrinkage that would result in the distor-tion of the type, or even the metal shrinkingaway from the mold wall. Since bismuth ex-pands during solidification, the molten metalcan actually expand to fill molds fully, therebyensuring accurate casting and repeatable type-faces.

5.43 Do you expect to see larger solidification shrink-age for a material with a bcc crystal structureor fcc? Explain.

The greater shrinkage would be expected fromthe material with the greater packing efficiencyor atomic packing factor (APF) in a solid state.Since the APF for fcc is 0.74 and for bcc it is0.68, one would expect a larger shrinkage fora material with a fcc structure. This can alsobeen seen from Fig. 3.2 on p. 84. Note, how-ever, that for an alloy, the answer is not as sim-ple, since it must be determined if the alloy-ing element can fit into interstitial positions orserves as a substitutional element.

5.44 Describe the drawbacks to having a riser thatis (a) too large, or (b) too small.

The main drawbacks to having a riser too largeare: the material in the riser is eventuallyscrapped and has to be recycled; the riser hassto be cut off, and a larger riser will cost more

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to machine; an excessively large riser slows so-lidification; the riser may interfere with solidifi-cation elsewhere in the casting; the extra metalmay cause buoyancy forces sufficient to sepa-rate the mold halves, unless they are properlyweighted or clamped. The drawbacks to hav-ing too small a riser are mainly associated withdefects in the casting, either due to insufficientfeeding of liquid to compensate for solidifica-tion shrinkage, or shrinkage pores because thesolidification front is not uniform.

5.45 If you were to incorporate lettering on a sandcasting, would you make the letters protrudefrom the surface or recess into the surface?What if the part were to be made by invest-ment casting?

In sand casting, where a pattern must be pre-pared and used, it is easier to produce lettersand numbers by machining them into the sur-face of a pattern; thus the pattern will have re-cessed letters. The sand mold will then haveprotruding letters, as long as the pattern isfaithfully reproduced. The final part will thenhave recessed letters.

In investment casting, the patterns are pro-duced through injection molding. It is easier toinclude recessed lettering in the injection mold-ing die (instead of machining protruding let-ters). Thus, the mold will have recessed let-ters and the pattern will have protruding let-ters. Since the pattern is a replica of the finalpart, the part will also have protruding letters.

In summary, it is generally easier to produce re-cessed letters in sand castings and protrudingletters in investment casting.

5.46 List and briefly explain the three mechanismsby which metals shrink during casting.

Metals shrink by:

(a) Thermal contraction in the liquid phasefrom superheat temperature to solidifica-tion temperature,

(b) Solidification shrinkage, and

(c) Thermal contraction in the solid phasefrom the solidification temperature toroom temperature.

5.47 Explain the significance of the “tree” in invest-ment casting.

The tree is important because it allows simul-taneous casting of several parts. Since signifi-cant labor is involved in the production of eachmold, this strategy of increasing the number ofparts that are poured per mold is critical to theeconomics of investment casting.

5.48 Sketch the microstructure you would expect fora slab cast through (a) continuous casting, (b)strip casting, and (c) melt spinning.

The microstructures are as follows:

Continuouscast

Strip cast Melt spunP

roce

ssin

gd

irect

ion

Note that the continuous cast structure showsthe columnar grains growing away from themold wall. The strip-cast metal has been hotrolled immediately after solidification, and isshown as quenched, prior it is annealed to ob-tain an equiaxed structure. The melt-spunstructure solidifies so rapidly that there are noclear grains (an amorphous metal).

5.49 The general design recommendations for a wellin sand casting are that (a) its diameter shouldbe twice the sprue exit diameter, and (b) thedepth should be approximately twice the depthof the runner. Explain the consequences of de-viating from these rules.

Refer to Figure 5.10 for terminology used in thisproblem.

(a) Regarding this rule, if the well diameter ismuch larger than twice the exit diameter,liquid will not fill the well and aspiration ofthe molten metal may result. On the otherhand, if the diameter is small compared tothe sprue exit diameter, and recognizingthat wells are generally not tapered, thenthere is a fear of aspiration within the well(see the discussion of sprue profile in Sec-tion 5.4 starting on p. 199.

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(b) If the well is not deeper than the runner,then turbulent metal first splashed intothe well is immediately fed into the cast-ing, leading to aspiration and associateddefects. If the well is much deeper, thenthe metal remains in the well and can so-lidify prematurely.

5.50 Describe the characteristics of thixocasting andrheocasting.

Thixocasting and rheocasting involve castingoperations where the alloy is in the slushy stage.Often, ultrasonic vibrations will be used to en-sure that dendrites remain in solution, so thatthe metal is a slurry of molten continuous phaseand suspended particles. In such casting op-erations, the molten metal has a lower super-heat and, therefore, requires less cycle time,and shrinkage defects and porosity can be de-creased. This is further described in Section5.10.6 starting on p. 233.

5.51 Sketch the temperature profile you would ex-pect for (a) continuous casting of a billet, (b)sand casting of a cube, (c) centrifugal castingof a pipe.

This would be an interesting finite-element as-signment if such software is made availableto the students. Consider continuous casting.The liquid portion has essentially a constanttemperature, as there is significant stirring ofthe liquid through the continuous addition ofmolten metal. The die walls extract heat, andthe coolant spray at the die exterior removesheat even more aggressively. Thus, a sketch ofthe isotherms in continuous casting would be asfollows:

MoldMolten metal

Isotherms

Conductionboundary

Convectionboundary

The cube and the pipe are left to be completedby the student.

5.52 What are the benefits and drawbacks to hav-ing a pouring temperature that is much higherthan the metal’s melting temperature? Whatare the advantages and disadvantages in hav-ing the pouring temperature remain close to themelting temperature?

If the pouring temperature is much higher thanthat of the mold temperature, there is less dan-ger that the metal will solidify in the mold,and it is likely that even intricate molds canbe fully filled. This situation makes runners,gates, wells, etc., easier to design because theircross sections are less critical for complete moldfilling. The main drawback is that there isan increased likelihood of shrinkage pores, coldshuts, and other defects associated with shrink-age. Also there is an increased likelihood of en-trained air since the viscosity of the metal willbe lower at the higher pouring temperature. Ifthe pouring temperature is close to the melt-ing temperature, there will be less likelihoodof shrinkage porosity and entrained air. How-ever, there is the danger of the molten metalsolidifying in a runner before the mold cavityis completely filled; this may be overcome withhigher injection pressures, but clearly has a costimplication.

5.53 What are the benefits and drawbacks to heatingthe mold in investment casting before pouringin the molten metal?

Heating the mold in investment casting is advis-able in order to reduce the chilling effect of themold, which otherwise could lead to low metalfluidity and the problems that accompany thiscondition. Molds are usually preheated to someextent. However, excessive heating will com-promise the strength of the mold, resulting inerosion and associated defects.

5.54 Can a chaplet also act as a chill? Explain.

A chaplet is used to position a core. It has ageometry that can either rest against a moldface or it can be inserted into a mold face. Ifthe chaplet is a thermally conductive material,it can also serve as a chill.

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5.55 Rank the casting processes described in thischapter in terms of their solidification rate.For example, which processes extract heat thefastest from a given volume of metal and whichis the slowest?

There is, as expected, some overlap between thevarious processes, and the rate of heat transfercan be modified whenever desired. However, ageneral ranking in terms of rate of heat extrac-tion is as follows: Die casting (cold chamber),squeeze casting, centrifugal casting, slush cast-

ing, die casting (hot chamber), permanent moldcasting, shell mold casting, investment casting,sand casting, lost foam, ceramic-mold casting,and plaster-mold casting.

5.56 The heavy regions of parts typically are placedin the drag in sand casting and not in the cope.Explain why.

A simple explanation is that if they were to beplaced in the cope, they would develop a highbuoyancy force that would tend to separate themold and thus develop flashes on the casting.

Problems

5.57 Referring to Fig. 5.3, estimate the followingquantities for a 20% Cu-80% Ni alloy: (1) liq-uidus temperature, (2) solidus temperature, (3)percentage of nickel in the liquid at 1400◦C(2550◦F), (4) the major phase at 1400◦C, and(5) the ratio of solid to liquid at 1400◦C.

We estimate the following quantities from Fig.5.3 on p. 192: (1) The liquidus temperature is1400◦C (2550◦F). (2) The solidus temperatureis 1372◦C (2500◦F). (3) At 2550◦F, the alloy isstill all liquid, thus the nickel concentration is80%. (4) The major phase at 1400◦C is liquid,with no solids present since the alloy is not be-low the liquidus temperature. (5) The ratio iszero, since no solid is present.

5.58 Determine the amount of gamma and alphaphases (see Fig. 5.4b) in a 10-kg, AISI 1060steel casting as it is being cooled to the follow-ing temperatures: (1) 750◦C, (2) 728◦C, and(3) 726◦C.

We determine the following quantities fromFig. 5.6 on p. 197: (a) At 750◦C, the alloy isjust in the single-phase austenite (gamma) re-gion, thus the percent gamma is 100% (10 kg),and alpha is 0%. (b) At 728◦C, the alloy isin the two-phase gamma-alpha field, and theweight percentages of each is found by the lever

rule (see Example 5.1):

%α =(

xγ − xo

xγ − xα

)× 100%

=(

0.77 − 0.600.77 − 0.022

)× 100%

= 23% or 2.3 kg

%γ =(

xo − xα

xγ − xα

)× 100%

=(

0.60 − 0.0220.77 − 0.022

)× 100%

= 77% or 7.7 kg

(c) At 726◦C, the alloy is in the two-phase al-pha and Fe3C field. No gamma phase is present.Again the lever rule is used to find the amountof alpha present:

%α =(

6.67 − 0.606.67 − 0.022

)×100% = 91% or 9.1 kg

5.59 A round casting is 0.3 m in diameter and 0.5 min length. Another casting of the same metalis elliptical in cross section, with a major-to-minor axis ratio of 3, and has the same lengthand cross sectional area as the round casting.Both pieces are cast under the same conditions.What is the difference in the solidification timesof the two castings?

66

Referring to Fig. 5.3, estimate the following quantities for a 20% Cu-80% Ni alloy: (1) liq-uidus temperature, 92) solidus temperature, (3) percentage of nickel in the liquid at 1673 K, (4) the major phase at 1673 K, and (5) the ratio of solid to liquid at 1673 K.

We estimate the following quantities from Fig. 5.3 on p. 192: (1) The liquidus temperature is 1673 K. (2) The solidus temperature is 1645 K. (3) At 1645 K, the alloy is still all liquid, thus the nickel concentration is 80%. (4) The major phase at 1673 K is liquid, with no solids present since the alloy is not below the liquidus temperature. (5) The ratio is zero, since no solid is present.

Determine the amount of gamma and alpha phases (see Fig. 5.4b) in a 10 kg, AISI 1060 steel casting as it is being cooled to the following temperatures: (1) 1023 K, (2) 1001 K, and(3) 999 K.

We determine the following quantities from Fig. 5.6 on p. 197: (a) At 1023 K, the alloy is just in the single-phase austenite (gamma) region, thus the percentage gamma is 100%(10 kg), and alpha is 0%. (b) At 1001 K, the alloy is in the two-phase gamma-alpha fi eld, and the weight percentages of each is found by the lever

(c) At 999 K, the alloy is in the two-phase al-

and cross-sectional area as the round casting.

For the same length and cross sectional area(thus the same volume), and the same castingconditions, the same C value in Eq. (5.11) onp. 205 on p. 205 should be applicable. The sur-face area and volume of the round casting is

Around = 2πrl + 2πr2 = 0.613 m2

Vround = πr2l = 0.0353 in2

Since the cross-sectional area of the ellipse isthe same as that for the cylinder, and it has amajor and minor diameter of a and b, respec-tively, where a = 3b, then

πab = πr2

3b2 = r2 → b =

√(0.15)2

3or b = 0.0866 m, so that a = 0.260 m. The sur-face area of the ellipse-based part is (see a ba-sic geometry text for the area equation deriva-tions):

Aellipse = 2πab + 2π√

a2 + b2l = 1.002 m2

The volume is still 0.0353 in2. According toEq. (5.11) on p. 205, we thus have

Tround

Tellipse=

(V/Around)2

(V/Aellipse)2=

(Aellipse

Around

)2

= 2.67

5.60 Derive Eq. (5.7).

We note that Eq. (5.5) on p. 200 gives a rela-tionship between height, h, and velocity, v, andEq. (5.6) on p. 201 gives a relationship betweenheight, h, and cross sectional area, A. With thereference plate at the top of the pouring basin(and denoted as subscript 0), the sprue top isdenoted as 1, and the bottom as 2. Note thath2 is numerically greater than h1. At the topof the sprue we have vo = 0 and ho = 0. As afirst approximation, assume that the pressurespo, p1 and p2 are equal and that the frictionalloss f is negligible. Thus, from Eq. (5.5) wehave

ho +po

ρg+

v2o

2g= h1 +

p1

ρg+

v21

2g+ f

or, solving for v1,

0 = h1 +v21

2g→ v1 =

√2gh1

Similarly,

ho +po

ρg+

v2o

2g= h2 +

p2

ρg+

v22

2g+ f

orv2 =

√2gh2

Substituting these results into the continuityequation given by Eq. (5.6), we have

A1v1 = A2v2

A1

√2gh1 = A2

√2gh2

A1

A2=

√2gh2√2gh1

=√

h2

h1

which is the desired relationship.

5.61 Two halves of a mold (cope and drag) areweighted down to keep them from separat-ing due to the pressure exerted by the moltenmetal (buoyancy). Consider a solid, spheri-cal steel casting, 9 in. in diameter, that is be-ing produced by sand casting. Each flask (seeFig. 5.10) is 20 in. by 20 in. and 15 in. deep. Theparting line is at the middle of the part. Esti-mate the clamping force required. Assume thatthe molten metal has a density of 500 lb/ft3 andthat the sand has a density of 100 lb/ft3,

The force exerted by the molten metal is theproduct of its cross-sectional area at the partingline and the pressure of the molten metal due tothe height of the sprue. Assume that the spruehas the same height as the cope, namely, 15 in.The pressure of the molten metal is the prod-uct of height and density. Assuming a densityfor the molten metal of 500 lb/ft3, the pres-sure at the parting line will be (500)(15/12) =625 lb/ft2, or 4.34 psi. The buoyancy force isthe product of projected area and pressure, or(625)(π)(9/12)2 = 1100 lb. The net volume ofthe sand in each flask is

V = (20)(20)(15) − (0.5)(

4π

3

)(9)3

or V = 4473 in3 = 2.59 ft3. For a sand den-sity of 100 lb/ft3, the cope weighs 454 lb. Un-der these circumstances, a clamping force of1100 − 259 ≈ 850 lb is required.

67

Vround = πr2l = 0.2277 cm2

Two halves of a mold (cope and drag) are weighted down to keep them from separating due to the pressure exerted by the molten metal (buoyancy). Consider a solid, spherical steel casting, 22.86 cm in diameter, that is being produced by sand casting. Each fl ask (see Fig. 5.10) is 50.8 cm by 50.8 cm and 38.1 cm deep. The parting line is at the middle of the part. Estimate the clamping force required. Assume that the molten metal has a density of 8009.2317 kg/m3 and that the sand has a density of 1601.8463 kg/m3.

The force exerted by the molten metal is the product of its cross-sectional area at the parting line and the pressure of the molten metal due to the height of the sprue. Assume that the sprue has the same height as the cope, namely, 38.1 cm. The pressure of the molten metal is the product of height and density. Assuming a density for the molten metal of8009.2317 kg/m3, the pressure at the parting line will be (8009.2317)(0.381) =3051.52 kg/m2, or 30 kPa. The buoyancy force is the product of projected area and pressure, or (3051.52)(π)(0.2286)2 = 500 kg. The net volume of the sand in each fl ask is

V = (50.8)(50.8)(38.1) – (0.5)

4π3

⎛⎝⎜

⎞⎠⎟

(22.86)3

or V = 73,300 cm3 = 0.0733 m3. For a sand density of 1601.8463 kg/m3, the cope weighs 206 kg. Under these circumstances, a clamping force of 1601.8463 – 206 = 855.85 kg is required.

For the same length and cross-sectional area (thus the same volume), and the same casting conditions, the same C value in Eq. (5.11) on p. 205 should be applicable. The surface area and volume of the round casting is

height, h, and cross-sectional area, A. With the

The volume is still 0.089662 cm2. According to Eq. (5.11) on p. 205, we thus have

5.62 Would the position of the parting line in Prob-lem 5.61 influence your answer? Explain.

The position of the parting line does have an in-fluence on the answer to Problem 5.54, because(a) the projected area of the molten metal willbe different and (b) the weight of the cope willalso be different.

5.63 Plot the clamping force in Problem 5.61 as afunction of increasing diameter of the casting,from 10 in. to 20 in.

Note in this problem that as the diameter ofthe casting increases, the cross-sectional areaof the molten metal increases, hence the buoy-ancy force also increases. At the same time,the weight of the cope decreases because of thelarger space taken up by the molten metal. Us-ing the same approach as in Problem 5.61, theweight of the casting as a function of diameteris given by

Fb = ρV = (500)(π

6d3

)=

(261 lb/ft3

)d3

The volume of sand in the cope is given by:

V = (20)(20)(15)(

112

)3

− (0.5)(π

6

)d3

= 3.47 ft3 − 0.261d3

Therefore, the weight of the sand is given by:

Fw = ρsandV

=(100 lb/ft3

) (3.47 ft3 − 0.261d3

)= 347 lb −

(26.1 lb/ft3

)d3

The required clamping force is given by equilib-rium as

Fc = Fb − Fw

=(261 lb/ft3

)d3 − 347 lb

+(26.1 lb/ft3

)d3

=(287 lb/ft3

)d3 − 347 lb

This equation is plotted below. Note that for asmall diameter, no clamping force is needed, asthe weight of the cope is sufficient to hold thecope and drag together.

5.64 Sketch a graph of specific volume vs. temper-ature for a metal that shrinks as it cools fromthe liquid state to room temperature. On thegraph, mark the area where shrinkage is com-pensated for by risers.

The graph is as follows. See also Fig. 5.1b onp. 189.

5.65 A round casting has the same dimensions asin Problem 5.59. Another casting of the samemetal is rectangular in cross-section, with awidth-to-thickness ratio of 3, and has the samelength and cross-sectional area as the roundcasting. Both pieces are cast under the sameconditions. What is the difference in the solid-ification times of the two castings?

The castings have the same length and cross-sectional area (thus the same volume) and thesame casting conditions, hence the same Cvalue. The total surface area of the round cast-ing, with l = 500 mm and r = 150 mm, is

Around = 2πrl + 2πr2

= 2π(150)(500) + 2π(150)2

= 6.13 × 105 mm2

68

Cla

mp

forc

e, k

g

680.4

453.6

226.8

0

–226.8

Casting diameter, cm12.7 25.4 38.1 50.80

from 25.4 cm to 50.8 cm.

Fb = ρV = 8009.2317π6

d3⎛⎝⎜

⎞⎠⎟

= (4193.6 kg/m3)d3

V = (50.8)(50.8)(38.1)

1100

⎛⎝⎜

⎞⎠⎟

3

– (0.5)π6

⎛⎝⎜

⎞⎠⎟

d3

= 0.0983 m3 – 0.261d3

Fw = ρsandV

= (1601.8463 kg/m3)(0.0983 m3 – 0.261d3)

= 157.5 kg – (418 kg/m3)d3

Fc = Fb – Fw

= (4193.6 kg/m3)d3 – 157.5 kg + (418 kg/m3)d3

= (4611.6 kg/m3)d3 – 157.5 kg

The cross-sectional area of the round casting isπr2 = π(150)2 = 70, 680 mm2. The rectangularcross section has sides x and 3x, so that

70, 680 = 3x2 → x = 153 mm

hence the perimeter of the rectangular castingwith the same cross-sectional area and axes ra-tio of 3 is 1228 mm. The total surface area is

Arect = 2(70, 680) + (1228)(500)

or Arect = 7.55 × 105 mm2. According toChvorinov’s rule, cooling time for a constantvolume is inversely proportional to surface areasquared. Therefore,

trecttround

=(

6.13 × 105

7.55 × 105

)2

= 0.66

5.66 A 75-mm thick square plate and a right circu-lar cylinder with a radius of 100 mm and heightof 50 mm each have the same volume. If eachis to be cast using a cylindrical riser, will eachpart require the same size riser to ensure properfeeding of the molten metal? Explain.

Recall that it is important for the riser to so-lidify after the casting has solidified. A castingthat solidifies rapidly would most likely requirea smaller riser than one which solidifies over alonger period of time. Lets now calculate therelative solidification times, using Chvorinov’srule given by Eq. (5.11) on p. 205 on p. 205.For the cylindrical part, we have

Vcylinder = πr2h = π(0.1 m)2(0.050 m)

or Vcylinder = 0.00157 m3. The surface area ofthe cylinder is

Acylinder = 2πr2 + 2πrh

= 2π(0.1)2 + 2π(0.1)(0.05)= 0.0942 m2

Thus, from Eq. (5.11) on p. 205 on p. 205,

tcylinder = C

(0.001570.0942

)2

=(2.78 × 10−4

)C

For a square plate with sides L and heighth = 0.075 m, and the same volume as the cylin-der, we have

Vplate = 0.00157 m3 = L2h = L2(0.075 m)

Solving for L yields L = 0.144 m. Therefore,

Aplate = 2L2 + 4Lh

= 2(0.144)2 + 4(0.144)(0.075)

or Aplate = 0.0847 m2. From Eq. (5.11) onp. 205 on p. 205,

tplate = C

(0.001570.0847

)2

=(3.43 × 10−4

)C

Therefore, the cylindrical casting will takelonger to solidify and will thus require a largerriser.

5.67 Assume that the top of a round sprue has a di-ameter of 4 in. and is at a height of 12 in. fromthe runner. Based on Eq. (5.7), plot the pro-file of the sprue diameter as a function of itsheight. Assume that the sprue has a diameterof 1 in. at its bottom.

From Eq. (5.7) on p. 201 and substituting forthe area, it can be shown that

d21

d2=

√h

h1

Therefore,

d =

√d21

√h1

h→ d = Ch−0.25

The difficulty here is that the reference locationfor height measurements is not known. Oftenchokes or wells are used to control flow, but thisproblem will be solved assuming that properflow is to be attained by considering hydrody-namics in the design of the sprue. The bound-ary conditions are that at h = ho, d = 4 (whereho is the height at the top of the sprue from thereference location) and at h = ho +12 in., d = 1in. The first boundary condition yields

4 = C(ho)−0.25 or C = 4h0.25o

The second boundary condition yields

1 = C(ho + 12)−0.25 =(4h0.25

o

)(ho + 12)−0.25

This equation is solved as ho = 0.047 in., sothat C = 1.863. These values are substitutedinto the expression above to obtain

d = 1.863(h + 0.047)−0.25

69

Assume that the top of a round sprue has a diameter of 10.16 cm and is at a height of30.48 cm from the runner. Based on Eq. (5.7), plot the profi le of the sprue diameter as a function of its height. Assume that the sprue has a diameter of 2.54 cm at its bottom.

ary conditions are that at h = ho, d = 10.16 cm (where ho is the height at the top of the sprue from the reference location) and at h = ho + 30.48 cm, d = 2.54 cm. The fi rst boundary condition yields

10.16 = C(ho)–0.25 or C = 10.16ho

0.25

The secondary boundary condition yields

2.54 = C(ho + 30.48)–0.25

= (10.16ho0.25)(ho + 30.48)–0.25

This equation is solved as ho = 0.11938 cm, so that C = 5.972. These values are substituted into the expression above to obtain

d = 5.972(h + 0.11938)–0.25

A 75 mm-thick square plate and a right circu-

longer period of time. Let’s now calculate the relative solidifi cation times, using Chvorinov’s rule given by Eq. (5.11) on p. 205. For the cylindrical part, we have

Thus, from Eq. (5.11) on p. 205

or Aplate = 0.0847 m2. From Eq. (5.11) on p. 205,

Note that ho is the location of the bottom ofthe sprue and that the sprue is axisymmetric.The sprue shape, based on this curve, is shownbelow.

5.68 Estimate the clamping force for a diecastingmachine in which the casting is rectangular,with projected dimensions of 75 mm x 150 mm.Would your answer depend on whether or notit is a hot-chamber or cold-chamber process?Explain.

The clamping force is needed to compensate forthe separating force developed when the metalis injected into the die. When the die is full,and the full pressure is developed, the separat-ing force is F = pA, where p is the pressure andA is the projected area of the casting. Note thatthe answer will depend on whether the opera-tion is hot- or cold-chamber, because pressuresare higher in the cold-chamber than in the hot-chamber process.

The projected area is 11,250 mm2. In the hot-chamber process, an average pressure is takenas 15 MPa (see Section 5.10.3), although thepressure can range up to 35 MPa. If we use anaverage pressure, the required clamping force is

Fhot = pA = (35)(11, 250) = 394 kN

For the cold-chamber process and using a mid-range pressure of 45 MPa, the force will be

Fcold = pA = (45)(11, 250) = 506 kN

5.69 When designing patterns for casting, pattern-makers use special rulers that automatically in-corporate solid shrinkage allowances into theirdesigns. Therefore, a 12-in. patternmaker’sruler is longer than a foot. How long should apatternmaker’s ruler be for the making of pat-terns for (1) aluminum castings (2) malleablecast iron and (3) high-manganese steel?

It was stated in Section 5.12.2 on p. 248 thattypical shrinkage allowances for metals are 1

8to 1

4 in./ft, so it is expected that the ruler bearound 12.125-12.25 in. long. Specific shrink-age allowances for these metals can be obtainedfrom the technical literature or the internet.For example, from Kalpakjian, ManufacturingProcesses for Engineering Materials, 3rd ed.,p. 280, we obtain the following:

ShrinkageMetal allowance, %Aluminum 1.3Malleable cast Iron 0.89High-manganese steel 2.6

From the following formula,

Lruler = Lo(1 + shrinkage)

We find that for aluminum,

LAl = (12)(1.013) = 12.156 in.

For malleable cast iron,

Liron = (12)(1.0089) = 12.107 in.

and for high-manganese steel,

Lsteel = (12)(1.026) = 12.312 in.

Note that high-manganese steel and malleablecast iron were selected for this problem becausethey have extremely high and low shrinkageallowances, respectively. The aluminum rulerfalls within the expected range, as do mostother metals.

5.70 The blank for the spool shown in the accompa-nying figure is to be sand cast out of A-319, analuminum casting alloy. Make a sketch of thewooden pattern for this part. Include all nec-essary allowances for shrinkage and machining.

70

2.54 cm

30.48 cm

10.16 cm

designs. Therefore, a 30.48 cm patternmaker’s

It was stated in Section 5.12.2 on p. 248 that typical shrinkage allowances for metals are 1.042 to 2.084 cm/m, so it is expected that the ruler be around 31.522–32.564 cm long. Specifi c shrinkage allowances for these metals can be obtained from the technical literature or the internet. For example, from Kalpakjian, Manufacturing Processes for Engineering Materials, 3rd ed., p. 280, we obtain the following:

LAl = (30.48)(1.013) = 30.876 cm

Liron = (30.48)(1.0089) = 30.75 cm

Lsteel = (30.48)(1.026) = 31.272 cm

0.50 in.

0.45 in.

4.00 in.

3.00 in.

The sketch for a typical green-sand casting pat-tern for the spool is shown below. A cross-sectional view is also provided to clearly in-dicate shrinkage and machining allowances, aswell as the draft angles. The important ele-ments of this pattern are as follows (dimensionsin inches):

(a) Two-piece pattern.(b) Locating pins will be needed in the pattern

plate to ensure that these features alignproperly.

(c) Shrinkage allowance = 5/32 in./ft.(d) Machining allowance = 1/16 in.(e) Draft = 3◦.

5.71 Repeat Problem 5.70, but assume that the alu-minum spool is to be cast using expendable-pattern casting. Explain the important differ-ences between the two patterns.

A sketch for a typical expandable-pattern cast-ing is shown below. A cross-sectional view isalso provided to clearly show the differencesbetween green-sand (from Problem 5.70) andevaporative-casting patterns. There will besome variations in the patterns produced bystudents depending on which dimensions are as-signed a machining allowance. The importantelements of this pattern are as follows (dimen-sions in inches):

(a) One-piece pattern, made of polystyrene.

(b) Shrinkage allowance = 5/32 in./ft

(c) Machining allowance = 1/16 in.

(d) No draft angles are necessary.

5.72 In sand casting, it is important that the copemold half be held down with sufficient force tokeep it from floating when the molten metalis poured in. For the casting shown in theaccompanying figure, calculate the minimumamount of weight necessary to keep the copefrom floating up as the molten metal is pouredin. (Hint: The buoyancy force exerted by themolten metal on the cope is related to the effec-tive height of the metal head above the cope.)

71

1.27 cm

1.143 cm

7.62 cm

10.16 cm

11.633 cm 3.81 cm

0.0524 rad (typical)

1.4224 cm

1.32 cm

10.287 cm 7.72

cm

(b) Shrinkage allowance = 1.3 cm/m

(c) Machining allowance = 0.15875 cm

(c) Shrinkage allowance = 1.3 cm/m

(d) Machining allowance = 0.15875 cm(e) Draft = 0.0524 rad

in centimeters):

sions in centimeters):

A A

Section A-A

3.00

2.002.50

1.000.50

2.00 1.00

1.00

2.50

4.00

5.00

R = 0.75

Material: Low-carbon steel Density: 0.26 lb/in3

All dimensions in inches

3.00

The cope mold half must be sufficiently heavyor be weighted sufficiently to keep it from float-ing when the molten metal is poured into themold. The buoyancy force, F , on the cope is ex-erted by the metallostatic pressure (caused bythe metal in the cope above the parting line)and can be calculated from the formula

F = pA

where p is the pressure at the parting line andA is the projected area of the mold cavity. Thepressure is

p = wh = (0.26 lb/in3)(3.00 in.) = 0.78 psi

The projected mold-cavity area can be calcu-lated from the dimensions given on cross sectionAA in the problem, and is found to be 10.13 in2.Thus, the force is

F = (0.78)(10.13) = 7.9 lb

5.73 The optimum shape of a riser is spherical toensure that it cools more slowly than the cast-ing it feeds. Spherically shaped risers, however,are difficult to cast. (1) Sketch the shape of ablind riser that is easy to mold, but also hasthe smallest possible surface area-to-volume ra-tio. (2) Compare the solidification time of theriser in part (a) to that of a riser shaped like aright circular cylinder. Assume that the volumeof each riser is the same, and that for each theheight is equal to the diameter (see Example5.2).

A sketch of a blind riser that is easy to cast isshown below, consisting of a cylindrical and ahemispherical portions.

Hemisphere

h=r

r

Note that the height of the cylindrical portionis equal to its radius (so that the total height ofthe riser is equal to its diameter). The volume,V , of this riser is

V = πr2h +12

(4πr3

3

)=

5πr3

3

Letting V be unity, we have

r =(

35π

)1/3

The surface area of the riser is

A = 2πrh + πr2 +12

(4πr2

)= 5πr2

Substituting for r, we obtain A = 5.21. There-fore, from Eq. (5.11) on p. 205, the solidificationtime, t, for the blind riser will be

t = C

(V

A

)2

= C

(1

5.21

)2

= 0.037C

72

1.272.54

5.086.35

7.62

2.54

6.35

10.16

12.7

R = 1.905 cm

Material: Low-carbon steelDensity: 7.196775 g/cm3

All dimensions in centimeters

2.545.08

7.62

AA in the problem, and is found to be 65.355 cm3. Thus, the force is

F = (5.5)(65.355) = 3.59 kg

p = wh = (7.196775 g/cm3)(7.62 cm) = 5.5 kPa

hemispherical portion.

From Example 5.2, we know that the solidifica-tion time for a cylinder with a height equal toits diameter is 0.033C. Thus, this blind riserwill cool a little slower, but not much so, and iseasier to cast.

5.74 The part shown in the accompanying figureis a hemispherical shell used as an acetabular(mushroom shaped) cup in a total hip replace-ment. Select a casting process for this partand provide a sketch of all patterns or toolingneeded if it is to be produced from a cobalt-chrome alloy.

This is an industrially-relevant problem, as thisis the casting used as acetabular cups for totalhip replacements. There are several possibleanswers to this question, depending on the stu-dent’s estimates of production rate and equip-ment costs. In practice, this part is producedthrough an investment casting operation, wherethe individual parts with runners are injectionmolded and then attached to a central sprue.The tooling that would be required include: (1)a mold for the injection molding of wax into thecup shape. (2) Templates for placement of thecup shape onto the sprue, in order to assureproper spacing for even, controlled cooling. (3)Machining fixtures. It should be noted that thewax pattern will be larger than the desired cast-ing, because of shrinkage as well as the incor-poration of a shrinkage allowance.

5.75 A cylinder with a height-to-diameter ratio ofunity solidifies in four minutes in a sand cast-ing operation. What is the solidification timeif the cylinder height is doubled? What is thetime if the diameter is doubled?

From Chvorinov’s rule, given by Eq. (5.11) on

p. 205, and assuming n = 2 gives

t = C

(V

A

)2

= C

[ (πd2h/4

)(πd2/2 + πdh)

]2

= C

(dh

2d + 4h

)=

4 min

Solving for C,

C = (4 min)(

2d + 4h

dh

)2

If the height is doubled, then we can use d2 = dand h2 = 2h to obtain

t = C

(d2h2

2d2 + 4h2

)2

= (4 min)(

2d + 4h

dh

)2 (d(2h)

2d + 4(2h)

)2

= (4 min)(

4d + 8h

2d + 8h

)2

If d = h, then

t = (4 min)(

12h

10h

)2

= 5.76 min

If the diameter is doubled, so that d3 = 2d andh3 = h, then

t = C

(d3h3

2d3 + 4h3

)2

= (4 min)(

2d + 4h

dh

)2 ((2d)(h)

2(2d) + 4(h)

)2

= (4 min)(

4d + 8h

4d + 4h

)2

or, for d = h,

t = (4 min)(

12h

8h

)2

= 9 min

5.76 Steel piping is to be produced by centrifugalcasting. The length is 12 feet, the diameter is 3ft, and the thickness is 0.5 in. Using basic equa-tions from dynamics and statics, determine therotational speed needed to have the centripetalforce be 70 times its weight.

73

Steel piping is to be produced by centrifugal casting. The length is 3.6576 m, the diameter is 0.9144 m, and the thickness is 1.27 cm. Using basic equations from dynamics and statics, determine the ratational speed needed to have the centripetal force be 70 times its weight.

The centripetal force can be obtained from anundergraduate dynamics textbook as

Fc =mv2

r

where m is the mass, v is the tangential veloc-ity, and r is the radius. It is desired to havethis force to be 70 times its weight, or

70 =Fc

W=

mv2/r

mg=

v2

rg

since r is the mean radius of the casting, or 1.25ft, v can be solved as

v =√

(70)rg =√

(70)(1.25)(32.2)

or v = 53 ft/sec or 637 in./sec. The rotationalspeed needed to obtain this velocity is

ω =v

r=

53 ft/sec1.25 ft

= 42.4 rad/sec

This is equivalent to 405 rev/min.

5.77 A sprue is 12 in. long and has a diameter of 5in. at the top, where the metal is poured. Themolten metal level in the pouring basin is takenas 3 in. from the top of the sprue for design pur-poses. If a flow rate of 40 in3/s is to be achieved,what should be the diameter of the bottom ofthe sprue? Will the sprue aspirate? Explain.

Assuming the flow is frictionless, the velocityof the molten metal at the bottom of the sprue(h = 12 in. = 1 ft) is

v =√

2gh =√

2(32.2)(1)

or v = 8.0 ft/s = 96 in./s. For a flow rate of 40in3/s, the area needs to be

A =Q

v=

40 in3/s96 in./s

= 0.417 in2

For a circular runner, the diameter would thenbe 0.73 in., or roughly 3

4 in. Compare this tothe diameter at the bottom of the sprue basedon Eq. (5.7), where h1 = 3 in., h2 = 15 in., andA1 = 19.6 in2. The diameter at the bottom ofthe sprue is calculated from:

A1

A2=

√h2

h1

A2 =A1√h2/h1

=19.6√15/3

= 8.8 in2

d =

√4π

A2 = 3.34 in

Thus, the sprue confines the flow more than isnecessary, and it will not aspirate.

5.78 Small amounts of slag often persist after skim-ming and are introduced into the molten metalflow in casting. Recognizing that the slag ismuch less dense than the metal, design moldfeatures that will remove small amounts of slagbefore the metal reaches the mold cavity.

There are several dross-trap designs in use infoundries. (A good discussion of trap designis given in J. Campbell, Castings, 1991, ReedEducational Publishers, pp. 53-55.) A conven-tional and effective dross trap is illustrated be-low:

It designed on the principle that a trap at theend of a runner will take the metal through therunner and keep it away from the gates. Thedesign shown is a wedge-type trap. Metal en-tering the runner contacts the wedge, and theleading front of the metal wave is chilled andattaches itself to the runner wall, and thus it iskept out of the mold cavity. The wedge must bedesigned so as to avoid reflected waves that oth-erwise would recirculate the dross or the slag.

The following design is a swirl trap:

74

since r is the mean radius of the casting, or 0.381 m, v can be solved as

v = (70)rg = (70)(0.381)(9.81)

or v = 16.17 m/s or 1615.44 cm/s. The rotational speed needed to obtain this velocity is

ω = vr

= 16.170.381

= 42.4 rad/s

This is equivalent to 405 rev/min.

A sprue is 30.48 cm long and has a diameter of 12.7 cm at the top, where the metal is poured. The molten metal level in the pouring basin is taken as 7.62 cm from the top of the sprue for design purposes. If a fl ow rate of 655.48 cm3/s is to be achieved, what should be the diameter of the bottom of the sprue? Will the sprue aspirate? Explain.

Assuming the fl ow is frictionless, the velocity of the molten metal at the bottom of the sprue (h = 30.48 cm) is

v = 2gh = 2(9.81)(30.48)

or v = 2.44 m/s = 244 cm/s. For a fl ow rate of 655.48 cm3/s, the area needs to be

A = Q

v= 655.48 cm3 /s

244 cm/s = 2.686 cm2

For a circular runner, the diameter would then be 1.854 cm, or roughly 1.905 cm. Compare this to the diameter at the bottom of the sprue based on Eq. (5.7), where h1 = 7.62 cm, h2 = 38.1 cm, and A1 = 126.45 cm2. The diameter at the bottom of the sprue is calculated from:

A1

A2

=h2

h1

A2 = A1

h2 h1

= 126.45

38.1 7.62 = 56.55 cm2

d =

4π

A2 = 8.49 cm

It is designed on the principle that a trap at the

This is based on the principle that the dross orslag is less dense than the metal. The metal en-ters the trap off of the center, inducing a swirl inthe molten metal as the trap fills with moltenmetal. Because it is far less dense than themetal, the dross or slag remains in the center ofthe swirl trap. Since the metal is tapped fromthe outside periphery, dross or slag is excludedfrom the casting.

5.79 Pure aluminum is being poured into a sandmold. The metal level in the pouring basin is10 in. above the metal level in the mold, andthe runner is circular with a 0.4 in. diameter.What is the velocity and rate of the flow of themetal into the mold? Is the flow turbulent orlaminar?

Equation (5.5) on p. 200 gives the metal flow.Assuming the pressure does not change appre-ciably in the channel and that there is no fric-tion in the sprue, the flow is

h1 +v21

2g= h2 +

v22

2g

Where the subscript 1 indicates the top of thesprue and 2 the bottom. If we assume that thevelocity at the top of the sprue is very low (aswould occur with the normal case of a pouringbasin on top of the sprue with a large cross-sectional area), then v1 = 0. The velocity atthe bottom of the sprue is

v22 = 2g(h1 − h2)

or

v2 =√

2g∆h =√

2(32.2 ft/s2)(12 in/ft)(10 in)

or v2 = 87.9 in./s. If the opening is 0.4-in. indiameter, the area is

A =π

4d2 =

π

4(0.4)2 = 0.126 in2

Therefore, the flow rate is

Q = v2A = (87.9)(0.126) = 11.0 in3/s.

Pure aluminum has a density of 2700 kg/m3

(see Table 3.3) and a viscosity of around 0.0015Ns/m2 around 700◦C. The Reynolds number,from Eq. (5.10) on p. 202, is then (using v =

87.9 in/s = 2.23 m/s and D = 0.4 in.=0.01016m),

Re =vDρ

η

=(2.23 m/s)(0.01016 m)(2700 kg/m3)

0.0015 Ns/m2

or Re= 40,782. As discussed in Section 5.4.1starting on p. 199, this situation would rep-resent turbulence, and the velocity and/or di-ameter should be decreased to bring Re below20,000 or so.

5.80 For the sprue described in Problem 5.79, whatrunner diameter is needed to ensure a Reynoldsnumber of 2000? How long will a 20 in3 castingtake to fill with such a runner?

Using the data given in Problem 5.79, aReynolds number of 2000 can be achieved byreducing the channel diameter, so that

Re = 2000 =vDρ

η=

(2.23)(2700)0.0015

D

or D = 0.000498 m = 0.0196 in.

For this diameter, the initial flow rate would be

Q = v2A =πv2

4D2 =

π

4(87.9)(0.0196)2

= 0.0266 in3/sec

This means that a 20 in3 casting would take 753s (about 12 min) to fill and only if the initialflow rate could be maintained, which is gener-ally not the case. Such a long filling time isnot acceptable, since it is likely that metal willsolidify in runners and thus not fill the moldcompletely. Also, with such small a small run-ner, additional mechanisms need to be consid-ered. For example, surface tension and fric-tion would severely reduce the velocity in theReynolds number calculation above.

This is generally the case with castings; to de-sign a sprue and runner system that maintainslaminar flow in the fluid would result in exces-sively long fill times.

5.81 How long would it take for the sprue in Problem5.79 to feed a casting with a square cross-sectionof 6 in. per side and a height of 4 in.? Assumethe sprue is frictionless.

75

25.4 cm above the metal level in the mold, and the runner is circular with a 1.016 cm diameter.

v2 = 2g∆h

= 2(9.81 m/s2 )(100 cm/m)(25.4 cm)

or v2 = 223.2 cm/s. If the opening is 1.016 cm in diameter, the area is

A = π4

d2 =

π4

(1.016)2 = 0.81 cm2

Therefore, the fl ow rate is

Q = v2A = (223.2)(0.81) = 180.955 cm3/s.

Pure aluminium has a density of 2700 kg/m3

(see Table 3.3) and a viscosity of around0.0015 Ns/m2 around 973 K. The Reynolds

number, from Eq. (5.10) on p. 202, is then (using v = 2.23 m/s and D = 0.01016 m),

or D = 0.000498 m.

For this diameter, the initial fl ow rate would be

Q = v2A =

πv2

4D2 = π

4(223.2)(0.000498)2

= 0.435 cm3/s

This means that a 327.74 cm3 casting would take753 s (about 12 min) to fi ll and only if the initial fl ow rate could be maintained, which is generally not the case. Such a long fi lling time is not acceptable, since it is likely that metal will solidify in runners and thus not fi ll the mold completely. Also, with such a small runner, additional mechanisms need to be considered. For example, surface tension and friction would severely reduce the velocity in the Reynolds number calculation above.

How long would it take for the sprue in Problem 5.79 to feed a casting with a square cross section of 15.24 cm per side and a height of 10.16 cm? Assume the sprue is frictionless.

For the sprue described in Problem 5.79, what runner diameter is needed to ensure a Reynolds number of 2000? How long will a 0.508 m3 casting take to fi ll with such a runner?

Note that the volume of the casting is 144in3, with a constant cross-sectional area of 36in2. The velocity will change as the mold fills,because the pouring basin height above themolten metal will decrease. The velocity willvary according to:

v = c√

2gh =√

2gh

The flow rate is given by

Q = vA =vπd2

4=

πd2√

2gh

4

The mold cavity fills at a rate of Q/(36 in2), or

dh

dt=

Q

A= −πd2

√2gh

4A

where the minus sign has been added so thath refers to the height difference between themetal level in the mold and the runner, whichdecreases with respect to time. Separating thevariables,

dh√h

= −πd2√

2g

4Adt

Integrating,

(2√

h)6in

10in= −πd2

√2g

4A(t)t

0

From this equation and using d=0.4 in. andA = 36 in2, t is found to be 14.7 s. As a compar-ison, using the flow rate calculated in Problem5.79, the mold would require approximately 13s to fill.

5.82 A rectangular mold with dimensions 100 mm ×200 mm × 400 mm is filled with aluminum withno superheat. Determine the final dimensionsof the part as it cools to room temperature. Re-peat the analysis for gray cast iron.

Note that the initial volume of the box is(0.100)(0.200)(0.400)=0.008 m3. From Table5.1 on p. 206, the volumetric contraction foraluminum is 6.6%. Therefore, the box volumewill be

V = (1 − 0.066)(0.008 m3) = 0.007472 m3

Assuming the box has the same aspect ratio asthe mold (1:2:4) and that warpage can be ig-nored, we can calculate the dimensions of the

box after solidification as 97.7 mm × 195.5 mm× 391 mm. From Table 3.3 on p. 106, the melt-ing point of aluminum is 660◦C, with a coef-ficient of thermal expansion of 23.6 µm/m◦C.Thus, the total strain in cooling from 660◦C toroom temperature (25◦C) is

ε = α∆t = (23.6µm/m◦C)(660◦C − 25◦C)

or ε = 0.0150. The final box dimensions aretherefore 96.2 × 192.5 × 385 mm.

For gray cast iron, the metal expands uponsolidification. Assuming the mold will allowfor expansion, the volume after solidification isgiven by

V = (1.025)(0.008 m3) = 0.0082 m3

If the box has the same aspect ratio as the ini-tial mold cavity, the dimensions after solidifica-tion will be 100.8 × 201.7 × 403.3 mm. Usingthe data for iron in Table 3.3, the melting pointis taken as 1537◦C and the coefficient of ther-mal expansion as 11.5 µm/m◦C. Therefore,

ε = α∆t = (11.5µm/m◦C)(1537◦C − 25◦C)

or ε = 0.0174. Hence, the final dimensions are99.0 × 198.1 × 396 mm. Note that even thoughthe cast iron had to cool off from a higher initialtemperature, the box of cast iron is much closerto the mold dimensions than the aluminum.

5.83 The constant C in Chvorinov’s rule is given as3 s/mm2 and is used to produce a cylindricalcasting with a diameter of 75 mm and a heightof 125 mm. Estimate the time for the cast-ing to fully solidify. The mold can be brokensafely when the solidified shell is at least 20 mm.Assuming the cylinder cools evenly, how muchtime must transpire after pouring the moltenmetal before the mold can be broken?

Note that for the cylinder

A = 2(π

4d2

)+ πdh

= 2[π

4(75)2

]+ π(75)(125)

= 38, 290 mm2

V =π

4d2h =

π

4(75)2(125) = 5.522 × 105 mm3

76

Note that the volume of the casting is 2360 cm3,with a constant cross-sectional area of232.26 cm2. The velocity will change as the mold fi lls, because the pouring basin height above the molten metal will decrease. The velocity will vary according to:

The mold cavity fi lls at a rate of Q/(232.26 cm2), or

2 h( )

25.4 cm

15.24 cm=

πd2 2g

4A(t)0

t

From this equation and using d = 1.016 cm and A = 232.26 cm2, t is found to be 14.7 s. As a comparison, using the fl ow rate calculated in Problem 5.79, the mold would require approximately 13 s to fi ll.

therefore 96.2 mm � 192.5 mm � 385 mm.

From Chvorinov’s rule given by Eq. (5.11) onp. 205,

t = C

(V

A

)2

= (3 s/mm2)(

5.522 × 105

38, 290

)2

or t = 624 s, or just over 10 min to solidify.

The second part of the problem is far more dif-ficult, and different answers can be obtaineddepending on the method of analysis. Thesolution is not as straightforward as it mayseem initially. For example, one could say thatthe 20 mm wall is 53.3% of the thickness, sothat 0.533(624)=333 s is needed. However, thiswould not be sufficient because an annular sec-tion at an outside radius has more material thanone closer to the center. It is thus reasonableand conservative to consider the time requiredfor the remaining cylinder to solidify. Usingh = 85 mm and d = 35 mm, the solidificationtime is found to be 21.8 s. Therefore, one stillhas to wait 602 s before the mold can be broken.

5.84 If an acceleration of 100 g is necessary to pro-duce a part in true centrifugal casting and thepart has an inner diameter of 10 in., a meanouter diameter of 14 in., and a length of 25 ft.,what rotational speed is needed?

The angular acceleration is given by α = ω2r.Recognizing that the largest force is experi-enced at the outside radius, this value for r isused in the calculation:

α = ω2r = 100 g = 3220 ft/s2

Therefore, solving for ω,

ω =√

α/r =√(

3220 ft/s2)

/(0.583 ft)

or ω = 74 rad/s = 710 rpm.

5.85 A jeweler wishes to produce twenty gold ringsin one investment-casting operation. The waxparts are attached to a wax central sprue ofa 0.5 in. diameter. The rings are located infour rows, each 0.5 in. from the other on thesprue. The rings require a 0.125-in. diameterand 0.5-in. long runner to the sprue. Estimatethe weight of gold needed to completely fill therings, runners, and sprues. The specific gravityof gold is 19.3.

The particular answer will depend on the geom-etry selected for a typical ring. Let’s approxi-mate a typical ring as a tube with dimensionsof 1 in. outer diameter, 5/8 in. inner diameter,and 3/8 in. width. The volume of each ring isthen 0.18 in3, and a total volume for 20 rings of3.6 in3. There are twenty runners to the sprue,so this volume component is

V = 20(π

4d2

)L = 20

(π

4(0.125 in.)2

)(0.5 in.)

or V = 0.123 in3. The central sprue has alength of 1.5 in., so that its volume is

V =π

4d2L =

π

4(0.5 in.)2(1.5 in.) = 0.29 in3

The total volume is then 4.0 in3, not includingthe metal in the pouring basin, if any. The spe-cific gravity of gold is 19.3, thus its density is19.3(62.4 lb/ft3) = 0.697 lb/in3. Therefore, thejeweler needs 2.79 lb. of gold.

5.86 Assume that you are asked to give a quiz to stu-dents on the contents of this chapter. Preparethree quantitative problems and three qualita-tive questions, and supply the answers.

By the student. This is a challenging, open-ended question that requires considerable focusand understanding on the part of the students,and has been found to be a very valuable home-work problem.

Design

5.87 Design test methods to determine the fluidity ofmetals in casting (see Section 5.4.2 starting on

p. 203). Make appropriate sketches and explainthe important features of each design.

77

If an acceleration of 100 g is necessary to produce a part in true centrifugal casting and the part has an inner diameter of 25.4 cm, a mean outer diameter of 35.56 cm, and a length of 762 cm, what rotational speed is needed?

α = ω2r = 100 g = 981.456 m/s2

ω = α / r = (981.456 m/s2 ) / (17.77 cm)

A jeweler wishes to produce twenty gold rings in one investment-casting operation. The wax parts are attached to a wax central sprue of a 12.7 cm diameter. The rings are located in four rows, each 12.7 cm from the other on the sprue. The rings require a 0.3175 cm diameter and 12.7 cm-long runner to the sprue. Estimate the weight of gold needed to completely fi ll the rings, runners, and sprues. The specifi c gravity of gold is 19.3.

The particular answer will depend on the geometry selected for a typical ring. Let’s approximate a typical ring as a tube withdimensions of 2.54 cm outer diameter,1.5875 cm inner diameter, and 0.9525 cm width. The volume of each ring is then2.95 cm3, and a total volume for 20 rings of59 cm3. There are twenty runners to the sprue, so this volume component is

V = 20π4

d2⎛⎝⎜

⎞⎠⎟

L

= 20π4

(0.3175 cm)2⎛⎝⎜

⎞⎠⎟

(1.27 cm)

or V = 2 cm3. The central sprue has a length of 3.81 cm, so that its volume is

V =

π4

d2L =

π4

(1.27 cm)2(3.81 cm) = 4.83 cm3.

The total volume is then 65.55 cm3, not including the metal in the pouring basin, if any. The specifi c gravity of gold is 19.3, thus its density is 19.3(1000 kg/m3) = 19,300 kg/m3. Therefore, the jeweler needs 1.265 kg of gold.

The second part of the problem is far more diffi cult, and different answers can be obtained depending on the method of analysis. The solution is not as straightforward as it may seem initially. For example, one could say that the 20 mm wall is 53.3% of the thickness, so that 0.533(624) = 333 s is needed. However, this would not be suffi cient because an annular section at an outside radius has more material than one closer to the center. It is thus reasonable and conservative to consider the time required for the remaining cylinder to solidify. Using h = 85 mm and d = 35 mm, the solidifi cation time is found to be 21.8 s. Therefore, one still has to wait 602 s before the mold can be broken.

By the student. The designs should allow somemethod of examining the ability of a metalto fill the mold. One example, taken fromKalpakjian and Schmid, Manufacturing Engi-neering and Technology, 5th ed, Prentice-Hall,2001, is shown below.

Pouring cup

SprueFluidity index

5.88 The accompanying figures indicate various de-fects and discontinuities in cast products. Re-view each one and offer design solutions to avoidthem.

(a)

Fracture

(b)

Riser

Gate

Casting

(c)

Sink mark

(d)

Cold tearing

By the student. Some examples are for (a) frac-ture is at stress raiser, a better design would uti-lize a more gradual filet radius; (b) fracture atthe gate indicates this runner section is too nar-row and thus it solidified first, hence the gateshould be larger.

5.89 Utilizing the equipment and materials avail-able in a typical kitchen, design an experiment

to reproduce results similar to those shown inFig. 5.12.

A simple experiment can be performed withmelted chocolate and a coffee cup. If a part-ing agent is sprayed into the cup, and moltenchocolate is poured, after a short while the stillmolten center portion can be poured from thecup, leaving a solidified shell. This effect canbe made more pronounced by using chilling thecups first.

5.90 Design a test method to measure the perme-ability of sand for sand casting.

Permeability suggests that there is a potentialfor material to penetrate somewhat into theporous mold material. The penetration can bemeasured through experimental setups, such asusing a standard-sized slug or shape of sand, ap-plying a known pressure to one side, and thenmeasuring the flow rate through the sand.

5.91 Describe the procedures that would be involvedin making a bronze statue. Which casting pro-cess or processes would be suitable? Why?

The answer depends on the size of the statue.A small statue (say 100 mm tall) can be diecast if the quantities desired are large enough,or it can be sand cast for fewer quantities. Thevery large statues such as those found in publicparks, which typically are on the order of 1 to 3m tall, are produced by first manufacturing orsculpting a blank from wax and then using theinvestment-casting process. Another option fora large casting is to carefully prepare a ceramicmold.

5.92 Porosity developed in the boss of a casting isillustrated in the accompanying figure. Showthat by simply repositioning the parting line ofthis casting, this problem can be eliminated.

Cope

Drag

Boss

Riser Part

Core

78

Note in the figure that the boss is at some dis-tance from the blind riser. Consequently, theboss can develop porosity (not shown in the fig-ure, but to be added by the instructor) becauseof a lack of supply of molten metal from theriser. The sketch below shows a repositionedparting line that would eliminate porosity inthe boss. Note in the illustration below that theboss can now be supplied with molten metal asit begins to solidify and shrink.

5.93 For the wheel illustrated in the accompanyingfigure, show how (a) riser placement, (b) coreplacement, (c) padding, and (d) chills may beused to help feed molten metal and eliminateporosity in the isolated hob boss.

Rim

Hub boss

Four different methods are shown below.

(a) Riser

(b) Core

(c) Pads

(d) Chills

5.94 In the figure below, the original casting designshown in (a) was changed to the design shownin (b). The casting is round, with a vertical axisof symmetry. As a functional part, what advan-tages do you think the new design has over theold one?

1 in.(25 mm)

1.5 in.(38 mm)

1 in.(25 mm)

1 in.(25 mm)

Ribs or brackets

(a)

(b)

By the student. There are several advantages,including that the part thickness is more uni-form, so that large shrinkage porosity is lesslikely, and the ribs will control warping due tothermal stresses as well as increasing joint stiff-ness.

5.95 An incorrect and a correct design for castingare shown, respectively, in the accompanyingfigure. Review the changes made and commenton their advantages.

Outside core

(a) incorrect

(b) correct

Outside core

By the student. The main advantage of thenew design is that it can be easily cast withoutthe need for an external core. The original partwould require two such cores, because the ge-ometry is such that it cannot be obtained in asand mold without cores.

79

5.96 Three sets of designs for die casting are shownin the accompanying figure. Note the changesmade to original die design (number 1 in eachcase) and comment on the reasons.

(1)

(2)

Parting line

(a)

(1) (2)

(b)

Parting line

Parting line

(3)

(1) (2)

(c)

By the student. There are many observations,usually with the intent of minimizing changesin section thickness, eliminating inclined sur-faces to simplify mold construction, and to ori-ent flanges so that they can be easily cast.

5.97 It is sometimes desirable to cool metals moreslowly than they would be if the molds weremaintained at room temperature. List and ex-plain the methods you would use to slow downthe cooling process.

There can be several approaches to this prob-lem, including:

• Heated molds will maintain temperatureshigher than room temperature, but willstill allow successful casting if the moldtemperature is below the melting temper-ature of the metal.

• The mold can be placed in a container;heat from the molten metal will then warmthe local environment above room temper-ature.

• The mold can be insulated to a greater ex-tent, so that its steady-state temperatureis higher (permanent-mold processes).

• The mold can be heated to a higher tem-perature.

• An exothermic jacket can be placedaround the molten metal.

• Radiation heat sources can be used to slowthe rate of heat loss by conduction.

5.98 Design an experiment to measure the constantsC and n in the Chvorinov’s Rule [Eq. (5.11)].

The following are some tests that could be con-sidered:

• The most straightforward tests involveproducing a number of molds with a familyof parts (such as spheres, cubes or cylin-ders with a fixed length-to-diameter ratio),pouring them, and then breaking the moldperiodically to observe if the metal has so-lidified. This inevitably results in spilledmolten metal and may therefore a difficulttest procedure to use.

• Students should consider designing moldsthat are enclosed but have a solidificationfront that terminates at an open riser; theycan then monitor the solidification timescan then be monitored, and then deter-mine fit Eq. (5.11) on p. 205 to their data.

• An alternative to solidifying the metal isto melt it within a mold specially designedfor such an experiment.

5.99 The part in the accompanying figure is to becast of 10% Sn bronze at the rate of 100 partsper month. To find an appropriate casting pro-cess, consider all the processes in this chapter,then reject those that are (a) technically in-admissible, (b) technically feasible but too ex-pensive for the purpose, and (c) identify the

80

most economical process. Write a rationale us-ing common-sense assumptions about productcost.

The answers could be somewhat subjective, be-cause the particular economics are affected by

company capabilities and practices. The follow-ing summary is reasonable suggestion:

Process Note Cost rationaleSand casting This is probably

best.Shell-mold casting (a)Lost Foam Need tooling to

make blanks. Toolow of productionrate to justify.

Plaster mold (a)Ceramic mold (b)Lost Wax Need to make

blanks. Too low ofproduction rate tojustify, unless rapidtooling is used.

Vacuum casting (b)Pressure casting (b)Die casting (b)Centrifugal casting (b)CZ Process (b)Notes: (a) technically inadmissible; (b) Too ex-pensive.

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25.4 cm

2.54 cm

10.16 cm

25.4 cm

Ra = 3.175 mµ

Ra = 1.524 mµ

1.143 � 0.127 cm

82